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Derivative of Inverse Function

  1. Sep 28, 2010 #1
    1. The problem statement, all variables and given/known data

    Find g' (-1/2), where g(x) is the inverse of f(x) = x3 / (x2 + 1)

    2. Relevant equations

    g'(x) = 1/f' [g(x)]

    3. The attempt at a solution

    Everything else I tried got really messy, but I feel like I'm going out on a limb by:

    1) y=x3 / (x2+1)

    2) y=x3x-2 / 1

    3) y=x

    4) g '(0.5)= 1/ (0.53 /(0.52+1) = 10

    I don't even know if I'm going about it the right way. Any help would be awesome. Thanks.
  2. jcsd
  3. Sep 28, 2010 #2
    You say in 1) that
    y = x^3 / (x^2 + 1)
    and in 3) that
    y = x.

    If you put in x = 1 you see that in 1) you get 1/2 and in 3) you get 1. Something must hence be wrong.
  4. Sep 28, 2010 #3
    Very true. I realize something is wrong. That was just my attempt.
  5. Sep 28, 2010 #4
    But . . . I could still use some help with this problem, obviously.
  6. Sep 28, 2010 #5


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    Line 2 is incorrect, it should be

    y=x3x-2 / (1+x-2),

    which actually just makes things more complicated. Just differentiate the expression on line 1 directly.
  7. Sep 28, 2010 #6
    It would help to figure out what g(-1/2) was equal to

    Then you could just plug stuff into this relationship: g'(x) = 1/f' [g(x)] -> g'(-1/2) = 1/f' [g(-1/2)]

    Since we know that g is the inverse of f, we should try to solve when f(a) = -1/2. Then we would know a = g(-1/2).

    So -1/2 = x^3 / (x^2 + 1) -> -1/2*(x^2 + 1) = x^3 -> 0 = x^3 + 1/2x^2 + ½ which clearly has a solution at -1. So we know g(-1/2) = -1 which means: g'(-1/2) = 1/f'(-1) and you can do the rest from there id bet.
  8. Sep 28, 2010 #7
    So I differentiate to get


    And then I dared to simplify it down to


    But what I don't understand from all of this is that if I do that, am I plugging this in for the g(x) for the 1/ f '[g(x)]? That would be like plugging f '[f '(x)] right? I'm sorry I'm just trying to understand how to solve these types of problems.
  9. Sep 28, 2010 #8
    As the guy above said, if f(c) = -1/2 then g(-1/2) = c. To find out what g(-1/2) is we need to find c, that is to solve the equation f(x) = -1/2.

    When this is done you can just plug everything into the formula.
  10. Sep 29, 2010 #9
    Surely if I have a function y=f(x), then the inverse is just x=g(y) and it's derivative is [itex]\frac{dx}{dy}[/itex] which is a function of y. So that for:


    we have [itex]x=f^{-1}(y)[/itex] and so the derivative of the inverse at [itex]y_0[/itex] can be written as:


    So now let [itex]y_0=-1/2[/tex]. Then solve:


    which has three solutions. The inverse and it's derivative are multi-valued right since it's going to be a cube root correct? Therefore, just find the three roots [itex] (x_1,x_2,x_3)[/itex] and for example, then one derivative of the inverse at y=-1/2 is:


    Try and follow the following Mathematica code which explicitly computes the inverse, then finds it's (multivalued) derivative at y=-1/2:

    Code (Text):

    f[x_] := x^3/(x^2 + 1);
    mysols = x /. Solve[f[x] == y, x]
    N[D[mysols, y] /. y -> -2^(-1)]
    Last edited: Sep 29, 2010
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